Unit -13 Bunkers and Silos

Eunkers and Silos

UNIT 13 BUNKERS AND SILOS Structure Introduction Objectives

Components of Bunkers Airy's Theory Janssen's Theory IS: Code Specifications 13.5.1 13.5.2 13.5.3 13.5.4 13.5.5

Calculation of Loads as Per IS Code Factors Increasing the Bin Loads Analysis of Bunkers Procedure for Design of Bunkers Design Problem

Design of Silo Summary Answers to SAQS

13.1 INTRODUCTION In the previous units, you have studied gantry girders, plate girder bridges, truss bridges, bearings. In this unit, you are going to learn the theory and design related to bunkers and silos. The bunkers are large size shallow bins to store grains, coal and cement. In bunkers, the plane of rupture intersects the free surface of the stored material. Generally, steel bunkers are used to store coal at power plants and loco-running sheds. Generally, these are square or rectangular shaped. The silos are the deep bins for storage. They are circular in shape. The plane of rupture intersects the opposite side of the container.

Objectives After studying this unit, you should be able to understand Airy's theory, understand Janssen's theory, know the components of bunkers and silos, design the bunkers, and design the silos.

13.2 COMPONENTS OF BUNKERS The sectional elevation and plan of the bunker are shown Figure 13.1. 1)

Main girder,

2)

Cross girder,

3) 4) 5) 6)

Beam, Sloping plates, Stiffeners, and Openings.

Steel Structures ,End

Naif7 Girder

Girder

4 1 -

Figure 13.1

Main Girders : The main girders are provided parallel to the longitudinal sides. These are supported on cross girders. Cross Girders : These are provided parallel to the width.

Sloping Plates : These are provided in the bottom portion of the bunker. The inclination is more than the angle of repose of the material for self cleaning. Openings : These are provided at the bottom of the bunkers. The size is 500 mm square. Stiffeners : These are provided with the inclined plates. At top, there are connected with the main -girder. At bottom, these are connected with the bottom plates.

SAQ 1 1)

What is a bunker?

2)

What is a silo?

3)

At what places, the steel bunkers are used?

4)

Differentiate between a bunker and

5)

What are the components of a bunker?

6)

Write a note on openings in bunkers.

7)

Write a note on stiffeners used in bunkers.

> silo?

Bunkers and Silos

13.3 AIRY'S THEORY By using this theory the horizontal pressure per unit length of periphery and po/s)tlon of plane of rupture can be determined. The Airy's theory is actually based on Coulomb's wedge theory of Earth Pressure. Consider a wedge ABC of unit thickness.

--

Figure 13.2

8 = inclination of plane of rupture with horizontal thickness of wedge

Let

ABC= 1 unit

W = weight of wedge R1 & R2 = reactions to BC and BA respectively Rn and Pn= normal reactions due to R1 & R2 respectively h = depth of Bunker

b = breadth of Bunker

Now

1 W=-xAC.AB.w

2

Resolving forces acting on wedge in vertical direction

R, = (w - p1 pn)/(p sin 8 + cos 8)

Resolving forces on wedge in horizontal direction Ph=@,cOs 8 =R,sin8 Rn =

'h

s i n e - pcos 8

From (13.2) and (13.3)

P,, =

w (sin 8 - p cos 8) (p + pl) sin 8 + (1 - pp') cos 8

Substituting value of w and simplifying wh2

P h ' X

2

(tan 8 - p) (p + pl) tan2 8 + (1 - ppl) tan 8

Steel Structures

or

wh2 P --x - 2

(tan 8 - p) ( p + pl) tan2 8 + (1 - p p l ) tan 8 where, u = tan 8 - p

v = ( p + pl) tan2 8 + (1 - p p l ) tan 8 For maximum value of P,

..

..

udv - vdu d8

-

tan 8 - p sec2 8 ( p + pl) tan2 8 + (1 - p p l ) tan 8 2 ( p + p l ) tan 8 sec2 8 + (1 - ppl) sec2 8

On further simplification tan2 0 - 2 p tan 0 = P - w l ) (P +

..

tan 8 = p +

Substituting the value of tan 8 in Eqn. (13.5) and simplifying

P h represents total horizontal force per unit length of the wall at depth h. The pressure per unit area

substituting p = tan (I and p1= tan (I1 in Eqn. (13.6) we have

Ph= wh If

cos (I 1 + dsin $ sec $' sin ((I+ (I1)

P, = vertical load carried by wall

Total load carried by the wall will be perimeterlines Pw. The maximum depth upto which the shallow bin acts as a bunker may be found as follows

I

'

Bunkers and Silos

Note: Eqn. (13.6) is applicable for maximum depth h = hmax as in Eqn. (13.7) when depth of bin 'h' is greater than hmax, the bin becomes a deep bin (silo) plane of rupture intersects opposite wall at C in case of a silo

:.

CD = (h - b tan 0)

:.

W = wbh - - wb.b tan 0 2

1

...(i) L

J

Similar to shallow bin from Eqn. (i)

P, ='

w (sin 0 - p cos 0) (p + pl) sin 0 + (1 - ppl) cos 0

Substituting the value of w and. simplifying

P, =

or

L

1

(p + pl) tan 0 + (1 - ppl)

L

Ph=

\

1

/

(p + pl) tan 0 + (1 - ppl)

dPn d0

For maximum value of Ph -= 0

.,

L

\

1

(p +

tan 0 + (1 - ppl)

-WJ

J-\ -

( p + pl) sec2 0

On further simplification

Substituting the value of tan0 and simplifying

The pressure Ph is given by

I 1

!

dP1, dh

-

Note: b may be taken as length of side adjacent to the wall on which the pressure is to be determined for

a rectangular silo. The vertical taken by wall is Pw = p'~,,.The total vertical load is perimeter x Pw.

Structures

Values of

CL and

for same comon materials have been given in Table 13.1

Table 13.1 Materials

S.No.

P

P'

1)

Wheat

0.466

0.443

2)

Maize

0.521

0.432

3)

Cement

0.316

0.554

4)

Bituminous coal

0.700

0.700

13.4 JANSSEN'S THEORY

-

Assumptions 1)

Most of the weight of the material stored in the bin is supported by friction between the material and the vertical wall.

2)

Weight transferred to the hopper bottom is very less. (Hence Rankine or Coulomb's lateral pressure theory cannot be applied.

3)

The vertical wall of the bin is subjected to vertical force and horizontal \ pressure.

.

Derivation Let

dH = thickness of the elementary layer considered H = depth from the top P, = vertical intensity of pressure acting at top of the layer

(P,

+ dP,) = verjical intensity of pressure acting on the bottom of the layer Ph = horizontal pressure

f = stress due to friction

v = unit weight of stored material A = CISarea of material stored

P = bin's interior perimeter

R = Alp = Hydraulic mean depth of the c/s

@' = angle of friction on the walls of bin @ = angle of internal friction of stored material consider the equilibrium of the vertical forces acting as shown in Figure.

P,.A+y.A.dH=(P,+dP,).A+f.P . d h or

PV.A+~.A.~H=(PV+~PV).A+(~~+P~)P~~

...(i)

Bunkers and Silos

*

J

~

Y-P

when h = 0 -

or

kx

P, = 0

=

J

:.~

logH

=H

- Cllk R

+ constant

C = log h -

PI,= p~(1 + e- "9

R where, zo = 1

vk

Figure 13.4

values of (1 - e - h / Z ~are ) given in Table 13.2.

Table 13.2: Values of (1 - e-h/q) as per IS:4995-1968 for Criteria for Design of ~einforcedConcrete Bins (Silos)

Steel Structures

The CISof rhe silos is generally a circle. The hoop tension on the wall = P,

D 2

-

[D = diameter of silo] Additionally the vertical wall will be subjected to. vertical pressure transferred due to friction P ,

Total vertical pressure = P , x Perimeter = P,.P

= (yAH - AP,)

- The value of k

The pressure ratio k lies between can be accurately found out experimentally.

Table 13.3: Angle of Wall Friction and Pressure Ratio S. No.

Angle of Wall Friction

Pressure Ratio

6

h

Material

1)

Granular materials with mean particle diameter 2 0.2 mm

2)

Powdery materials (except wheat flour) with mean particle diameter < 0.06 mm

3)

Wheat flour

While Filing

While Emptying

While Filling

While Emptying

0.75@

0.6@

0.5

1 .O

1.O@

1.O@

0.5

0.7

SAQ 2 1)

What is the basis of Airy's theory?

2)

What are the assumptions made in Janssen's theory?

3)

What are the forces acting on the walls of a bin?

Bunkers and Silos

13.5 IS: CODE SPECIFICATIONS IS: 4995 (Part-I)-1974, gives the general requirements and assessment of bin loads. The parameters which influence the design of bunkers are: 1)

Unit weight of materials.

2)

Angle of internal friction.

3)

Angle of wall friction.

4)

Pressure ratio.

Table 13.4 gives the unit weight and angle of internal friction ((I) of some important stored materials.

Table 13.4 Material

-

Unit Weight &N/m3)

Apgle of Internal Friction (+)I

Degrees

Wheat

8.50

28

Paddy

5.75

36

Rice

9.00

33

Maize

8.00

30

Barley

6.90

27

Corn

8.00

27

Sugar

8.20

35

Wheat flour

7.00

30

Coal

8.00

35

Coke

4.30

30

Ash

6.50

30

Cement

15.50

25

Lime

16.50

25

Bin Loads

Figure 13.5: Uin Loads

Steel Structures

Three types of loads are caused by a stored material in a bin. These are

bh)acting on the

1)

Horizontal pressure

side walls.

2)

Vertical pressure @), acting on the cross-sectional area

3)

Frictional wall pressure @).,

'

the bill filling,

Governing loading cases are given in Table 13.5:

-[-T Table 13.5

Granular Material

PR~SIV

I

Emptying

Filling = Emptying

I

Ph

Filling

PV

--

filling = Emptying Filling .

-

13.5.1 Calculation of Loads as Per IS Code Case 1: Granular Materials

1)

Maximum pressure Table 13.6 -

, -

Name of Pressure

During Filling

T-

During

YR

Frictional wall pressure

I

Horizontal pressure

I Vertical pressure where,

y = unit weight of material stored, A . R = - ratlo

P

pf = coefficient of wall friction during filling, pe = coefficient of wall friction during emptying, Af= pressure ratio during filling, he = pressure ratio during emptying. 2)

Variation of Pressure along the Depth

where, p stands for pressure and suffix i stands for w, h or v.

Q o.-t9hZ

(a]

Hhirhever ;SLBSS

Reductton due

L

11

Bunkers and Silos

R During filling Zof= ~fIf R During emptying = -

he Pe

To reduce the load effect of the bin bottom, the horizontal pressure during emptying may be reduced upto a height of 1.2d or 0.75h (Figure 13.6) whichever is smaller from the bin bottom.

/

Case 2 : Powdery Materials Maximum design pressures are the same as in case (1). The lateral and vertical pressures

.

ph =pv = 0.6 yZ - during homogenization ph = 0.8 yzn where,

- during rapid

filling

Zn= (V- v0) t v = actual filling speed, (mh) vO= the min-filling speed (mh)

t = time lapse (hours) The values of vo for some material are given in Table 13.7 Table 13.7

Lime

) Wheat flour

~1

1.4 4.8

1I

During pneumatic emptying air under pressure is blown inside the bin through a number of small holes located in the bin walls near the bin bottom. This causes the liquefaction of the material in the lower portion of the bin and gives rise to higher values of p, and p, (Figure 13.7).

W

Figure 13.7: Pressure Scheme for Pneumatic Emptying

Steel Structures

13.5.2 Factors Increasing the Bin Loads 1)

Eccentric emptying.

2)

Arching of stored material.

3)

Discharge promoting devices.

4)

Aeration of stored material.

1) Eccentric Emptying Eccentric emptying of a bin gives rise to increased horizontal loads non-uniformly distributed over the periphery and extending over the full height of the bin. Eccentric outlets in bins shall be avoided as far as possible and where they have to be provided to meet functional requirements, due consideration shall be given in design to the increased pressure experienced by the walls. This increased pressure shall be considered for the purpose of design, to be acting both on the wall nearer to the outlet as well as on the wall on the opposite side. The enlarged shape of the bin which is required for the purpose of computation of the pressure Phi shall be obtained as shown in Figures 13.8 and 13.9. P'ha

out Lat

I

I

I

r c - ~ n b r b d Sri L,-------l I

Figure 13.8: Rectangular Bin

p'h cas cp

~ c t u a lBin

- Eccentric

Figure 13.9: Circular Bin

The effect of eccentric outlets may be ignored in design if the eccentricity is less than dl6 or the height of the bin is not greater than 2d, where d is the maximum possible diameter of the circle that can be inscribed in the bin.

Bunkers and Silos

2) Arching of Stored Material Some stored materials are susceptible to arching action across the bin walls. The frequent collapse of such order gives rise to increased vertical pressures. The vertical pressures on the bottom of the bin storing such materials shall be taken as twice the filling pressure, PVihowever the loadneed not be assumed to be greater than W.Z. 3) Discharge Promoting Devices Modern bins storing various materials may be provided with various discharge promoting devices such as inserts, bridge like structures above the outlet or relief nose. In all such cases the effective cross-section of the bin is locally reduced. Recent research has given an indication that in such bins, the horizontal wall pressures are excessively increased locally or along the entire bin height. 4) Aeration of Stored Material

When bins are provided with equipment for ventilating the bin filling at rest, a distinction must be made between b i n s for granular material and bins for powdery material. When the material is granular, an increase in the horizontal pressures is to be expected., Therefore, the horizontal pressure Ph (for filling) has to be increased by the inlet pressure of the air over that portion of the height of the bin in which the air inlets are located. From the level of the highest inlet upwards, this increase in pressure may be tappered off uniformly down to zero at the top of the bin. For powdery materials the measurements made so far do not indicate any significant increases in load when ventilating. Bins for storage of powdery materials are often equipped with devices for pneumatic emptying, and these bting about a loosening of the bin filling in the region of the outlet. In this case also, no significant increases in load due to the air supply have far been detected.

13.5.3 Analysis of Bunkers Let us consider a symmetrical rectangular bunker with trough bottom. For the analysis of loads, unit length is considered. The cross-section is showing in Figure 13.10.

Figure 13.10

Caculate the horizontal pressures at different points. ~ i n the d total pressure Phi , Ph and Ph,as shown in Figure 13.11. 2

Steel Structures

By taking moments about 0,

By taking moments about B,

Figure 13.11

The forces acting on the sloping sides are calculated as follows: Let P, and PC are the normal pressure at B and C, then the normal load acting on the sloping side at the centroid of the pressure diagram.

Figure 13.12

Normal pressure,

Tangential pressure,

\ " I where,

w, = self weight of hopper

a = angle of the hopper with the horizontal.

/ I

13.5.4 Procedure for Design of Bunkers Step 1: Force analysis

a)

Calculate the vertical forces.

b)

Calculate the horizontal forces using code specification.

c)

Calculate the bursting forces H , , H 2 , H3 and H4. Using equation of equilibrium.

I

d)

Calculate the pressure p, , p, , ph on trough walls.

I

e)

Calculate the normal and tangential pressures.

f)

Calculate the normal load on trough.

I

I i

I

Step 2: Design of trough plates

i

a)

Span = spacing of stiffeners.

b)

Considering truss-way bending, calculate the maximum bending,

where,

p = maximum normal pressure

L = span of trough plate.

i

c)

Calculate the thickness required

Min. thickness = 6 mm. i

Step 3: Design of inclined stiffeners in trough

I

I b

I I

i I

I

I

1

a)

Calcular the maximum BM and (M2) and direct tension at mid-span.

b)

Choose suitable T-section with plate.

c)

Calculate A, Ixx and ZxF

d) Check for tensile stress and bending stress. Step 4: Design of plate stgeners for trough These are provided perpendicular to the T-stiffeners. a)

Calculate the maximum BM.

b)

Calculate the section modulers Zrequired.

c)

Assuming thickness ( t ) , find the depth of plate

Step 5: Design of vertical plate

a)

'

P L2 Calculate the maximum BM, M - 4-2~12

Bunkers and Silos

Structures

b)

~ a l c u l i t e= t , =

dx

(Jbc

L

.

But min t = 6 mm

Step 6: Design of vertical stifleners

a)

pL2 Calculate the max. BM, M - 5-2x8

b)

M5 Calculate Zrequired =(Jbc

c)

Choose a standard T-section with plate

d)

Calculate A, Ixx, Zxx

e)

Check for bending stress.

Step 7: Design main (longitudinal) girder

a)

H1 L2 Calculate the moment due to Hl at top, M6 = 8

b)

-- M6 Calculate Zrequired '=Jbc

c)

Choose the suitable section.

Step 8: Design of horizontal girder

a)

Calculate the moment due to H3

b)

Calculate Zrequired.

c)

Select the suitable section.

13.5.5

Design Problem

Design a rectangular bunker 16 m length and 8 m width supported on ten columns. It stores maize. Height of vertical portion = 4 m. Height of hopper= 4 m .

Step 1: Force Analysis

$ for maize

= 30'

Unit weight

= 8.00 k ~ / m ~

For filling,

$ ; = 0.75 $ = 22.5'

For emptying, Pressure ratio, For filling, hf = 0.50 For emptying, he = 1.00 Ffor filling, P ; =tan

# =tan $ = 0.414

Ffor emptying, pj = tan 4: = 0.325 Taking 1 m length Weight wl=8x4x3.5=112kN Weight w 2 = 8 ~ [ $ x I ~ 3 ~ 5 ) = 5 6 i N Weight w3 = 8 x (0.5x 8) = 32 kN Horizontal forces

.

h is more (Emptying) zoe

(1 - e- ,,'$) = 0.39

(1

- e- 'f/%)

= 0.62

Horizontal pressure

(from table)

(from table) P,, =

* Cb

x 0.62

= 40.75 kN/m2

Total Horizontal Pressures

Bursting Forces on Walls

Figure 13.15

Figure 13.14

Taking moments about 0,

= 201.307

:.

H3 = 50.33 kN.

Taking moments about B,

H4 = 183.09 kN Ph + P h , + P h , = H 1 + H 2 - H 3 + H 4

:.

Sum of horizontal forces is zero.

Pressures on Trough Walls

At B , p v = 103.19 x 0.39 = 40.24 k ~ / r n ~ Vertical pressure due to weight on wall, p , = 8 x 4 = 32 k ~ / r n ~

'.' Pv > Pw :.

pv i s taken.

Vertical pressure at C due to weight of material = 8 x 8 = 64 kP!/rn2

At B,

vertical pressure = 40.24 kN/m2 horizontal pressure = 25.63 kN/m2

At C ,

vertical pressure = 64.00 kN/m2 horizontal pressure = 40.75 kN/m2 5.32 m

Length, BC = -= 4 rana=-=a=48.8 3.5 sin a = 7.5

sin 2a = 0.975

cos a = 0.65

Normal Pressures

Tangential Pressures At BY P ~ = B

(Pv - ~ 2

h )

sin 2 a

= 11.33 kN/m2

Normal load on trough wall

I

Figure 13.16

Step 2: Design of trough plates Adopt a spacing of 600 mm for stiffeners. The plates are bending in two-direction.

Bunkers and Silos

Steel Structures

Maximum BM, M =

Thickness,

& 2 x 12

.\1 6M, --

oh

c,

~ - . \ 1 6 X 0 . 7 5 ~ 1 0 =6.74mm 165 x 600

Adopt 8 mm thick plates. Step 3: Design of inclined stifeners in a trough

Length of the stiffeners = 5.32 mm length BM at the mid spacelm length

For 600 mm spacing, M2 = 0.6 x 143.88 = 86.33 kN-m

Let us try ISST 250 @ 375 Nlm with 350 mm wide plate. Thickness of plate is 10 mm For ISST 250 @ 375 Nlm

Figure 13.17

> 'required

..

Safe.

Step 4: Design of plate stifSeners in trough

Spacing of plate stiffeners = 600 mm (let)

= 2727.3 mm.

Adopting 10 mm thick plate,

.:

d = 40.5 mm.

Step 5: Design of vertical plates pL2 Maximum BM, M4 = -24

Thickness t =

Adopt 6 mm thickwall. Step 6: Design of vertical stiffeners

At A,

Ph=O

At B, ph = 25.63 kN/m2 Length, L = 4 m pL2 Max. BM, M -5-2x8

- 25'63 42 = 25.63 W-m 2x8 Try for ISST 200 @ 284 N/m with 280 mm wide x 10 mm thick plate. Properties of ISST 200 are: A = 3622 mm2

Cxx= 47.8 mm

Bunkers and Silos

A

Steel Structures

= 116.34mm

1,,= 1267.5x lo4+ 3622 (116.34 - 47.Q2

+ 28012lo3+ 280 x 10 x (116.34 - 205)~ = 5172.3 x lo4 mm4

xxx=

172'3 lo4= 444.6 103 mm3 116.34

-

ZSOhrn

4aonm

Figure 13.18

= 155.33x lo3 mm3

Zxx > ZreFire,. Hence safe.

Step 7: Design of Main Beam Length of the beam = 4 m H, = 17.09 kN/m

Bending moments M6 = 171)9x42 = 34.1g id.l-m 8

Adop ISLB 225 B 235 Nlm

(Zxx = 222.4 x lo3 mm3)

Step 8: Design of horizontal beam Length,

L =4 m H3 = 50.33 kN/m

= 610 x lo3 mm3

Use ISLB 350 O 495 Nlm (Zxx = 75 19 x lo3 mm3)

nunkers and Silos

)

13.6 DESIGN OF SILO Generally, the Silos are circular in shape. These are designed similar to bunkers. Design Procedure Step 1: Calculation of horizontal pressure

By using the coda1 provisions, find the horizontal and verucal pressures at different depths at some intervals say 3 m, 4 m15 m. Step 2: Calculation of max. hoop tension D HOOPtension, H, = (ph)max . 2 Step 3: Design of wall plate

Calculate total vertical load, self weight, weight of lining, weight of top cover. Calculate the vertical load. Calculate thickness of plate from combined loading. Step 4: Design of hopper

Calculate the total vertical load. Calculate the direct tension. Calculate the thickness =

Direct tension oarx 1000 mm

Step 5: Design of ring beam

Calculate the weight of stored material, self weight of silos lining cover, platform. Calculate the reaction, SF, BM, torsion and compression. Calculate

oac

9

oat, c u l l

obc

%c, cal

Check for combined stresses.

Example 13.1 Design a circular steel silo of 10 m height and 4 m internal diameter to store cement of unit weight 15.50 kN/m3 and Q = 25'.

Solution Step 1: Calculation of pressures

The mean size of particle is less than 0.06 mm. For powdery material, For filling,

@'f=1.0 @ = 25"

For emptying, Q'e = 1.O Q = 25O

I

I

I

Pressure ratio, For filling,

hf

For emptying,

A, = 0.7

Angle of wall friction,

= 0.5

Steel Structures

p,1 = tan$;

For filling,

= 0.47 For e m ~ t v i n ~ .

u l e = 0.47

0.6m

Figure 13.19 1

h .. 0

is more (emptying)

zoe

Horizontal pressure,

x 0.96

ph =

Pf

- 1 x 15.5 0.96 = 31.66m/m2 0.47 Vertical pressure, p, =

Pf

LfS

x 0.96 =

,= 0.50

= 63.32 kN/m2

Step 2: Max. Hoop Tension

D 4 H,=ph-=31.66x-=61.32kN/m 2 2 Step 3: Design of Wall Plate 7T

Total vertical load = - x 4* x 10 x 15.5= 1948 kN 4 Assume wt of silo

+ stiffeners and lining = 2 k ~ l r n ~

Bunkers and Silos

Total wt. = 2 x n x 4 x 10=251kN Weight of top cover = 4 kN/m2 n Total wt. = - ~ 4 ~ ~ 4 = 5 0 k N 4 Total vertical load = 1948 + 251 Vertical loadmm length =

+ 50 = 2249 kN

2249 x l d = 179 N/mm n x 4000

H, = 61.32 kN/m = 61.32 N/mm Let t be the thickness, 0.3 be the poisson ratio. 179 (0.3 x 61.32) Max. compressive stress = = 150 t t +

Adopt 8 mm thick plate. Since the thickness provide is very lower than the required, nominal stiffeners are provided @ 1.50 m spacing. Stiffeners ISA @ 6060, 6 mm Step 4: Design of hopper

Vertical load = - d 2 p, 4

= 795.70 kN

Weight of material is hopper

Self wt.

= 60 kN (let)

Total load

= 795.70

Length of hopper = Loadmm run

+ 194.68 + 60 =

J32= 3-45m

=

1050.38x lo3 = 84 N/mm n x 4000

Direct tension Assuming 8 mm thick plate tensile stress

t

1050.38 kN

..

Safe.

Steel Structures

Step 5: Design of Ring Beam Weight of material stored

Self wt. of silo lining, cover, platform = 251

Total load = 2143

+ 50 + 50

+ 351

= 2494 kN

= 2500 kN (say) Using 8 No. of supports. Reaction

=2500 - 312.5 kN 8

Shear force = 2500 - 156.25 kN 16 Bending moment at support

= 0.00827 WR = 0.00827 x 2500 x 2 = 41.35 kN-m.

2500 x 3 = 2206 kN Direct compression = - x 2 1.7 Assume oat = 1200 ~ / m r n ~ Gross area required =

2206 lo' = 18382.4 rnm2 120

Adopting the sect shown in Figure 13.20.

Figure 13.20

Bunkers and Silos

Length between two adjacent columns

I=-=nD nx4x1000=1570.8mm

8 8 1 1570.8 - 24.2 Slenderness ratio, A = -= -rnlin 64.7 From Table 5.1, oar= 146 ~/rnrn'

Hence safe.

SAQ 2 1)

What are the factors causing increase in load on bunkers?

2)

Design a bunker of size 12 m length x 6 width. It has 4 m depth vertical plate and height of trough is 4 m. Use coal for storing.

3)

Design a silo for time having 4 m dia and 16 m height. Assume any suitable data.

13.7 SUMMARY In this unit you have studied about the basic theories and design philosophy of bunkers and silos. After reading this unit, you can understand the theories related to bunkers and silos. The design procedures are given for easy understanding. After studying, you are able to design the steel bunkers and silos.

13.8 ANSWERS TO SAQs SAQ 1 1)

Section 13.1

2)

Section 13.1

Steel Structures

3)

Section 13.1

4)

Section 13.1

5

Section 13.2

6)

Section 13.2

7)

Section 13.2

SAQ 2 1)

Sub-section 13.5.2

2)

Sub-section 13.5.5

3)

Sub-section 13.6.2